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Prove the two groups are isomorphic. I may use the FHT. $\mathbb Z_2$ and $\mathbb Z_2 \times \mathbb Z_2/K$ where $K=\{(0,0),(0,1)\}$.

I do not need to prove there's a homomorphism so I found that there are two cosets, i.e $K$ and $K+1=\{(1,1),(1,0)\}$. I am trying to determine how to write a kernel for some function $f: \mathbb Z_2 \times \mathbb Z_2 \rightarrow \mathbb Z_2$. I would typically a have permutation-like set up comparing the two group and where zeros were in the group compared to the quotient group I would let them be the kernel.

Similar to: $\mathbb Z_2$ and $\mathbb Z_{6}/<2>$: $\bigl(\begin{smallmatrix} 0 & 1 & 2 & 3 &4 & 5 \\ 0 & 1 & 0 & 1 & 0 & 1 \end{smallmatrix}\bigr)$ thus the ker$(f)=\{0,2,4\}=<2>$ so theres a homomorphism from $\mathbb Z_6/<2>$ to $\mathbb Z_2$ and it follows from the FHT that $\mathbb Z_2$ is isomorphic to $\mathbb Z_6/<2>$.

So I'm not sure how to set this up. I know that kernel will have 2 elements and thus be generated by K.

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Both are groups with two elements. Upto isomorphism there is only one such group.

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  • $\begingroup$ So is enough of a proof to state that since $\mathbb Z_2$ and K both have 2 elements, they are isomorphic? $\endgroup$ – K Math Oct 25 '17 at 13:48
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    $\begingroup$ $\mathbb Z_2$ and $K$ are also isomorphic, but that is not the question. $(\mathbb Z_2 \times \mathbb Z_2)/K$ has two elements and is thus isomorphic to $\mathbb Z_2$. $\endgroup$ – MooS Oct 25 '17 at 13:51
  • $\begingroup$ I apologize this is what I meant to ask. But is there more to show? I didn't think having the same order was enough of a proof. $\endgroup$ – K Math Oct 25 '17 at 13:55
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    $\begingroup$ For me, there is nothing more to show. $\endgroup$ – MooS Oct 25 '17 at 13:55
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Take $f: \mathbb Z_2 \times \mathbb Z_2 \rightarrow \mathbb Z_2$ defined by $f(x,y)=x$. Then $f$ is surjective and $\ker f = K$.

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